Given two standard normal random RVs, X and Y, how do you find $P(X^2 + Y^2 \leq 1)$?
I approached this by integrating the bivariate using polar coordinates as $x^2 + y^2 = 1$. Not sure if this is correct way or if there's a better way.
My approach:
$$ P(X^2 + Y^2 \leq 1) = \frac{1}{2\pi}\int_{0}^{2\pi} \int_{0}^1 \exp(-\frac{1}{2}) dr d\theta \\ = \frac{1}{2\pi}\int_{0}^{2\pi} \exp(-\frac{1}{2}) d\theta \\ = \frac{1}{2\pi} \cdot 2\pi \exp(-\frac{1}{2}) \\ = \exp(-\frac{1}{2}) $$
Since $dxdy=rdrd\theta$ we get $$P(X^{2}+Y^{2} \leq 1)$$ $$=\frac 1 {2\pi} \int_0^{2\pi} \int _0^{1} e^{-r^{2}/2} rdrd\theta$$ $$ ={-e^{-r^{2}/2}}|_0^{1}=1-e^{-1/2}.$$